5t)8
Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
Section 5.4 Exponential Functions" Differentiation and lintegration
l. elnx = 4
x=4
9.
2. eln2x = 12
2x = 12
X
X=6
3. ex = 12
x = In 12 ,~ 2.485
In 84 = 2
x = 21n84 ~ 8.862
5000
-2
1 + e2x
4. 4ex = 83
ex = 83
4
x=
800 - 50
100 - ex/2
800
- 100 - e42
50
84 = e42
ln(8.~] ~ 3.033
5. 9 - 2ex = 7
2ex = 2
5000
- 1 + e2x
2
2499 = e2x
In 2499 = 2x
1
x =-ln 2499 ~ 3.912
2
ex =1
11. ~ x = 2
x = e2 ~ 7.389
X=0
6. -6 + 3ex = 8
3ex = 14
ex = 14
12, ln x2 = 10
X2 = e1o
x = ±e5 ~ ±148.413
3
~ 1.540
13o ln(x- 3) = 2
x-3=e2
7. 50e-x = 30
e-X = ~_
x = 3 +e2 ~ 10.389
5
14. In 4x = 1
4x = el
e
X ~ --
~ 0.511
3
40
.,/-~x + 2
=el=e
x+2
= e2
x
0.648
0.680
15. in x/-~x + 2=1
8. 200e-4"~ = 15
e-4X _ 15 _
200
4
e
= e2 -2 = 5.389
16o ln(x-2)2 = 12
(x- 2)2 = e’=
x-2=e6
x=2
+ e6 ~
405.429
© 2010 Brooks/Cole, Cengage Learning
Section 5.4 Exponential Functions." Differentiation and Integration 51)9
22. y = e-4z
17o y = e-~"
18. y = 1$ex
23. (a)
4"
Horizontal shift 2 units to the right
(b)
t
4
19o y=eX +2
-3
A reflection in the x-axis and a vertical shrink
(c)
1-3 -2 -1
123
-4
8
-1
/
Vertical shift 3 units upward and a reflection in the
y-axis
3
24. (a)
2
1
-3 -2 -1
-1
123
-8 --’-~ ~ ~ 10
-3"
-2
Horizontal asymptotes: y = 0 and y = 8
21, y = e-x2
Symmetric with respect to the y-axis
Horizontal asymptote: y = 0
(b)
l0
-8 ~ 10
-2
Horizontal asymptote: y = 4
25. y = Ce~
Horizontal asymptote: y = 0
Matches (c)
© 2010 Brooks/Cole, Cengage Learning
516
Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
26. y = Ce-~’~
31. f(x)
Horizontal asymptote: y = 0
g(x) = ln(x + 1)
Reflection in the y-axis
Matches (d)
42-
Vertical shift C units
Reflection in both the x- and y-axes
Matches (a)
~
246
C
l+e-ox
C
lim --C
x~o~l + e-~
c
lim
-0
x-~l + e-~
Horizontal asymptotes: y = C and y = 0
28. y-
32° f(x) = ex-’
g(x) = l+lnx
Matches (b)
29. f(x) : e2x
I
2
3
4
-|.
g(x) : ln-,/-~ = l2lnx
33.
g
-1
As x --~ 0% the graph off approaches the graph ofg.
lim.1 +
30. f(x) : ex/3
= e°’5
34. Using the result from Exercise 33:
g(x) = lnx3 = 31nx
lim 1+
y
= e~forr > O.
86-
35.
4-
/1’°°°’°°°
1,O0~,O00J
e ~ 2.718281828
2-2
~ 2.718280469
2
4
6
8
e > (1 +
-2-
36. 1 + 1 + ~ + -g + + ]~ + 7-T6 +
1,oo6,ooo)Too,°°°
1
= 2.71825396
e ~ 2.718281828
1
1 +~5040
1
e>l+l+½+++ +1-T6+7-~
-~4
© 2010 Brooks/Cole, Cengage Learning
Section 5.4 Exponential Functions." Differentiation and Integration 511
y : e3x
37. (a)
45. y = ex In x
y’ = 3e3x
y’(O) = 3
Tangent line: y - 1 = 3(x - 0)
y =3x+l
y = e-3x
y’ = _3e-3x
y’(O) = -3
Tangent line: y - 1 = -3(x - 0)
y = -3x + 1
y
38. (a)
= e2x
y’ = 2e2x
46. y = xex
y’ = xex + eX(1) = eX(x + l)
47. y = x3ex
,’: x3ex + 3x~(e~) : x~e~(x + 3): eX(x~ + 3x~)
48. y = x2e-x
y’= x2(-e-x) + 2xe-x : xe-’~(2 - x)
49. g(t) = (e-t + e’)3
g’(,) : 3(e-’ + et)2(e’- e-’)
y’(O) :
Tangent line: y - 1 = 2(x - 0)
y = 2x+l
50. g(t) : e-3/t2
g,(t) = e_3/t2(6t_3)= 6
t3 e3/t2
y = e-2x
(b)
y’ = _2e-2x
y’(o) --
51.
Tangent line: y - 1 = -2(x - 0)
y = -2x + 1
S(x):
52.
f’(x) : 2e2x
40. y = e-Sx
dY = _5e-SX
y = ln(1 + e2x)
dy -- 2e2x
dx 1 + e2x
l+ eX) = In(l+ eX)- In(l-ex)
ha(
Y= ~,~)
dy -- ex
dx 1 + ex
53.
41. y = e"/~
y-
ex
1 - ex
2
_ 2(eX +e_X)-~
eX + e-x
dY_dx 2(ex + e-X)-2(eX -e-x) = (e:~
dy e47
2ex
1 - e2x
+ e-:~)2
ex _ e-X
54.
42. y = e-x2
dY = _2xe-X2
43. y
= ex-4
y’
= ex-4
y ~~
dy
--
dx
--
2
ex + e-x
2
-2ex
44. f(x)= 3el-x2
f’(x) : 3ei-X2 (-2x) : -6xel-x2
© 2010 Brooks/Cole, Cengage Learning
512
Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
65.
--
2e2x
y = x2e~ -2xex +2ex, (1, e)
y’ = x2ex + 2xex - 2xex - 2ex + 2ex
= x2ex
y’(1) = e
Tangent line: y - e = e(x - 1)
57.
y
~
eX(sin x + cos x)
e’~(cos x -sin x) + (sin x + cos x)(e*)
66.
eX(2 cos x) = 2e~ cos x
58.
y = xex -ex, (1,0)
y’ = xex + ex _ ex = xex
y’O) = e
y = ln eX = x
Tangent line: y - 0 = e(x - 1)
y=ex-e
dx
67. :(x) = e-x ln x, (1,0)
59. F(x)= ::Xcose’dt
f’(x) = e-X(~;-e-Xlnx = e-~(-~-
F’(4 = ~os(e’~X). x± = ~os(~/
x
1’(1) =
Tangent line: y - 0 = e-l(x - 1)
11
60. F(x) = ![2x ln(t + 1) dt
y=
F’(x) = ln(e2x + 1)2e2~"
61.
:(x) = e’-’, (l, 0
e
e
68. f(x) = e~lnx, (1,0)
f’(x) =-e’-x, f’(1)=-I
:,(x)
= :x
s’(1) = e3
Tangent line: y - 1 = -l(x - 1)
y=-x+2
Tangent line: y - 0 = e3(x - 1)
62, y = e-2x+x2, (2, 1)
y = e3(x -1)
y’= (2x- 2)e-zr+xz, y’(2) = 2
Tangent line: y - 1 = 2(x - 2)
xey-’:
dx
y = 2x-3
y= ln(e:3)= x~,
63.
y’ = 2x
y’(-2) = -4
x + e-x
64. y : ~ e
2
’
=
(o, o)
ex _ e-x]
+ ey - 10 + 3-’: = 0
dx
dY(xeY
+ 3)= 10- ey
dx~
dy--10 - ey
dx xey + 3
(-2, 4)
Tangent line: y - 4 = -4(x + 2)
y = -4x - 4
xey - l Ox + 3 y = 0
69.
70.
eX~, + x2 _ y2 = 10
+ 2x-2ydy = 0
dx
dY(xe~Y - 2y)= -yex,’ - 2x
dx~
dy -- ye"~’ + 2x
dx
xex* - 2 y
y’(0) : 0
Tangent line: y = 0
© 2010 Brooks/Cole, Cengage Learning
Section 5.4 Exponential Functions." Differentiation and Integration 513
71.
xey + yex = 1, (0, 1)
xeYy’ + ey + yex + y’ex = 0
74. g(x) = ~x + eX ln x
At(0,1): e+l+y’ = 0
y’ = -e - 1
Tangent line: y - 1 = (-e - 1)(x - 0)
y : (-e-1)x + l
72.
l+ln(v) = ex-y, (1,1)
1,
[1 y’]
-~-[xy + y] = ex-y -
at (1, 1): [y’ + 1] = t- y’
ex + ex lnx
g’(x) = ~1 + --x
g"(X) =
1
4X3/2
1
4 x ./ x
Xex -- ex
ex
+ -- 2+ -- + eX ln x
X
X
eX(Zx
-1)
+
2
x
+eXlnx
y = 4e-x
y’ = _4e-x
y" = 4e-x
y"- y = 4e-x -4e-x = 0
y’ = 0
Tangent line: y - 1 = 0(x - 1)
y=l
73. f(x) = (3 + 2x)e-3x
y = e3x + e-3x
y’ = 3e3x _ 3e-3X
y" = 9e3x + 9e-3x
/,- = ÷ 9(e , ÷ _-o
f’(x) = (3 + 2x)(-3e-3x) + 2e-3x = (-7 - 6x)e-3x
f"(x) = (-7 - 6x)(-3e-3x) - 6e-3" = 3(6x + S)e-3x
77.
y = eX(cosxi~x + sin x/~x)
y’= eX(-.,4~ sin x/~x + ~ cos x/~x)+ eX(cos ~/-~x + sin w/-~x)
= eX[(1 + x/~) cos w/-~x + (1 - ~) sin ~x]
,"= eX[-(~ + 2)sin ~x+ (~- 2)cos ~x]+ e’;[(1 + ~)cos ~x+ (1- ~)sin ~x~
= eX[(-,- 2~) sin ~x + (-1 + 2~) cos ~x~
-2y’+ 3y= -2e’~(1 + ~)cos ~x+ (1- ~)sin ~x~+ 3eX[cos ~x + sin ~x~
= eX[O- 2 )cos (,+ )sin
Therefore, -2y’ + 3y = -y" ~ y" - 2y’ + 3y = 0.
78.
y = eX(3 cos 2x - 4 sin 2x)
y’ = eX(-6 sin 2x - 8cos2x) + eX(3 cos2x - 4sin 2x) = e’~(-lOsin 2x - 5 cos2x) = -SeX(2 sin 2x + cos 2x)
y" = -SEX(4 cos 2x - 2 sin 2x) - 5e’~(2 sin 2x + cos 2x) = -SEX(5 cos 2x) = -25ex cos 2x
y"- 2y’ = -25excos 2x - 2(-5eX)(2 sin 2x + cos 2x) = -5eX(3 cos 2x - 4 sin 2x) = -Sy
Therefore, y" - 2y’ = -5y ~ y" - 2y’ + 5y = 0.
© 2010 Brooks/Cole, Cengage Learning
514
Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
f’(x~ =
~x _ e-x
2
g(x) = ~l~g-(x-3)2/2
",/2Z
g’(x) = -1
--~z(x- 3)e_(X_3)2/2
= 0 when x = 0.
g"(x) 1= ~(~2)(x -4)e-(x-3)2/2
- eX + e-x >0
f ()-"’x"
2
Relative minimum: (0, 1)
t3,0.39
Relative maximum: 3,
Points of inflection:
(2,-~ze ),l _1/~’) (4,1_1_&e_,/21~, (2,0.242),(4,0.242)~/2;,r )
if(x)
ex + e-x
2
>0
o
ex _ e-X
f"(x} = ~ = 0 when x = 0.
2
Point of inflection: (0, 0)
83.
f(x) .= x2e-x
f~’(X) :--X2~°-x -t-
2xe-x
= xe-*(2 -x) = 0when x = 0,2.
f"(x) = -e-X(2x - x2) q- e-X(2 - 2x)
= e-X(x2 -4x+2)= 0whenx = 2_+
Relative minimum: (0, 0)
81. g(x) = -~ e-(x-2)2 /2
g’(x) = -1
--~z(x- 2)e_(x_2)2/2
1)(x -3)e-(’~-2)2/2 ~’
g,,(~) =1 ~;(~Relative maximum:
2, (2,0.399)
(~ff)
Points of inflection:
1 -1/2")
1 -1/2"]
-~ffe ),
3,-~ffe ) ~ (1,0.242),(3,0.242)
Relative maximum: (2, 4e-2)
x = 2_+~/~
Points of inflection:
/
~ (3.414, 0.384),(0.586, 0.191)
ll,
o
© 2010 Brooks/Cole, Cengage Learning
Section 5.4 Exponential Functions." Differentiation and Integration
84. f(x) = xe-~
f’(x) = -xe-x + e-~
= e-’~(1- x) = 0when x = 1.
87.
A = (base)(height)= 2xe-x2
dA
_4x2e_.~2 + 2e_.~2
dx
= 2e-~2(1- 2x2) = 0when x -
s"(4 : + (-e-DO- x)
= e-"(x-2) = 0whenx = 2.
A = x/~e-112
Relative maximum: (1, e-l)
3-
Point of inflection: (2, 2e-2)
2-
2
(1, e-I)
-2
2
-2
-1
/
-1-
88. (a)
85. g(t) : 1 + (2 + t)e-’
S(c)
= f(c + x)
-c
l Oce = 10(c + x)e@÷’)
g’(t) : -(1 + t)e-’
g"(t) : te-t
c
ce
Relative maximum: (-1,1 + e)~ (-1,3.718)
Point of inflection: (0, 3)
ec
c+x
--
eC+X
= (C + X)ec
cex =CWX
Cex --C = X
5
C --
86. f(x) = -2 + e3X(4- 2x)
f’(x) : e3X(-2)+ 3eaX(4 - 2x)
= e3’~10-6x = 0whenx =
X
10x2 x/(1-ex)
(C) A(X) x.... e"
e -1
()
(2.118, 4.591)
f"(x) : e3X(-6) + 3e3X(10 - 6x)
= e3X(24- 18x) = 0when x 3"= 4
Relative maximum: (-}, 96.942)
Point of inflection: (-}, 70.798)
100 (--53’96’942)
9
0
The maximum area is 4.591 for x = 2.118 and
f(x) = 2.547.
x
(d) c- e~-I
lim c =1 l~1
x--~. 0+
-0,5
o
’~-~J 2.5
lim c = 0
x---~ oo
0
4
0
Answers will vary. Sample answer:
As x approaches 0 from the right, the height of the
rectangle approaches 1.
As x approaches ~o, the height of the rectangle
approaches 0.
© 2010 Brooks/Cole, Cengage Learning
516
Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
89° f(x)
91. V = 15,000e-°’6286t, 0
= e2x
s’(x) : 2e2x
_< t _< 10
(a) 20.000
Let (x, y) = (x, e2X)be the point on the graphwhere the
tangent line passes through the origin. Equating slopes,
2e2X _ e2x - 0
x-O
1
2 =-
(b) d.._~V = _9429e_O.6286t
dt
x
When t = 1, -dV~ -5028.84.
dt
dV
When t = 5,- ~, -406.89.
dt
x = 15’ y = e, y’ = 2e.
P°int: (~, e/
C) 20,000
Tangent line: y- e = 2e(x-½1
y = 2ex
92. 1.56e-°’22t cos 4.9t _< 0.25 (3 inches equals one-fourth
foot.) Using a graphing utility or Newton’s Method, you
have t > 7.79 seconds.
o
2
90° Let (Xo, Y0) be the desired point on y = e-x.
y’ = -e-x
(Slope of tangent line)
1
= ex
y’
(Slope of normal line)
93.
h
0
5
10
15
20
y-e-Xo = eXO (x- Xo)
P
10,332
5583
2376
1240
517
You want (0, O) to satisfy the equation:
In P
9.243
8.627
7.773
7.123
6.248
_e-Xo = -xoeXO
1 = XOe2xO
12
a)
XOe2xO -- 1 = 0
Using a graphing utility or Newton’s Method,the
solution is x0 = 0.4263.
(0.4263, e-0"4263)
o
y = -0.1499h + 9.3018 is the regression line for
data (h, In P).
(b) lnP = ah +b
P = eah+b = ebeah
P = Ceah, C = eb
~x
123
a = -0.1499 and C = e9’3018 = 10,957.7.
So, P = 10,957.7e-°’~499h
(C) ~ 2,ooo
o
o
22
Section 5.4 Exponential Functions." Differentiation and Integration 5117
f(x) = ex/:,
f(O) = 1
f’(x) : e42,
f’(O) :7
For h = 5,-- = -776.3.
dh
f"(x) : e42,
f"(O) = -~
For h = 18,-dP~ -110.6.
dh
P~(x) = 1 +
dP = (10,957.71)(_0.1499)e_0.1499h
(d) -~
= -1642.56e-°’1499h
94. (a) Linear model: V = -1686.8t + 27501
Quadratic model: V = 109.52tz - 3001.1t + 31006
96.
,
½(x- o) : x7 + 1,
Pl(O) : 1
: 7’1
= -71
1 0)+ 0)21 P~(O) 1
P~(x) : 1 +-~(x- ~(x-
:
25,000
X2 X
= --+--+1
82
1
1
P~’(x) : 4
Quadratic
L
2,1.
1
-x + 7’ : -7
Linear[ 110
0
(b) The slope represents the average loss in value per year.
(c) Exponential model:
V = 30,582.68(0.90724)t = 30,582.68 e-0’09735t
(d) As t -~ 0, V --~ 0 for the exponential model. The
value tends to zero
(e) When t = 4, V’ = -2017 dollars/year
The values of f, P~, P~ and their first derivatives agree
at x = 0. The values of the second derivatives off and
P2 agree at x = 0.
7
When t = 8, V’ ~ -1366 dollars/year
95. f(x)= ex
f’(x) = e>:
f"(x) = e,<
f(O) = 1
f’(O) = 1
f"(O) = 1
Pl(x) = 1 +l(x-O)= 1 + x
P~(x) = 1 + l(xi O)=
1)(x- 0)2 = 1 + x +2
8
97. n = 12.12! = 12 . 11 ¯ 10... 3 . 2 ¯ 1 = 479,001,600
Stirlings Formula:
12! ~
,~~,, ~ 475,687,487
98° n = 15.15! = 15 . 14... 3 ¯ 2 . 1 = 1,307,674,368,000
Stirlings Formula:
15t ~
,~,,,, ~ 1,300,430,722,200
~ 1.3004 x 10l~
The values off, P~, and P2 and their first derivatives
agree at x = 0.
99. Let u = 5x, du = 5 dr.
[eSX(5"~ dr = e5x + C
111)0. Let u = -x4, d// = i4x3dr.
le-X4(_4~t ~x _-e-~4 ÷~
101. Let u = 2x-l, du = 2dr.
l
eII2x-1
= -~ e
2x-l( dr
) l_2x-1
2 dr = -~e
+C
© 2010 Brooks/Cole, Cengage Learning
518
Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
1t)2. Let u = 1 - 3x, du = -3 dx.
Ie,-3.~dx = _½Iel-a,(_3)dx = --~
112. Let u = e~ +e-x,du = (e~ - e-x) dx.
+C
(eX + e-X)2
103. Let u = x3,du = 3x2 dx.
-2
- --+C
x
-x
= 1 ~x3
e +e
104. Letu = ex +l, du = exdx.
ieX(e, + 1)2dx = l(eX + 1)2(eX)dx _ (e" + 1)~
--+C
3
1
105. Letu = ~,du =~&.
2x
x+1
114. Jfe + 2e
x
e
2 Ie"/~
e"G
= ex +2x-e-x + C
dx = 2e"/; + C
11~. le-X tan(e-X) dx : -I[tan(e-X)l(-e-X) dx
106. Let u =1
-- du = -2 dx.
2
X
~
= In cos(e-x) + C
7
gl/X2
116.’ " "~ln{ez~-~) dx ="’J{2x - l)dx
=x2-x+C
l~7. Let u = 1 +
1
1
117. ~o
[~e-2x dx = -+ f2 e-2~(-2)dx = [--e-2~l
L 2 Jo
~-x
e~__- 1
= -~(1 - e-2)- 2e2
4= -e-~ + 1
e3-x d~ = ~-e3-X~
k -13
e
108. Let u = 1 + e2~, du = 2e~-x dx.
e2x
1¯ f 2e2x
1 ln[1
+c
l~dx = -~al + e2-----Tdx =-2k ~+ eZX)
109. Let u = 1-ex,du =-exdx.
l~-~,]i- ~x dx =-1(1-eX)’/2(-e~)dx
ex _ e-X
I~ dx = ln(ex + e-x) + C
111. Letu = ex-ex,- du = (ex +e-x) dx.
e~ + ~-x
dx = lnex-e-x +C
x
!
1
11
119. 12 xe-x= dx = --~I~e-~2(-2x)dx
= ,2L ]o
=-+Ee-1 -1~
_ ~ - O/e) _ e -
1
2e
2
= --}(1 - ex)3/2 + C
110. Letu = e" +e-x,du = (ex - e-x) dx.
= 1 - --
120. l~2 x2ex3/2 dx
=. Z fO ~x3/2(2X21 dx
3 ~-~
\2 ~
_ 2IeX3/2~°
-3
~-~
=--~I1-~41- 2(e4 -3e4
1)
e _ e-X
© 2010 Brooks/Cole, Cengage Learning
,
Section 5.4 Exponential Fidnctions: Differentiation and Integration 519
3
3
121. Let u = --, du = --~ dx.
127. Let Id = ax2, dld = 2ax dx. (Assume a : 0.)
x
3 e3/x
y
1 3e3/X¢_Z~dx
= Ixear2 dx
_- 1--~a
f ax2
l__~_eaX2
ae (2ax) dx =2a
+C
_x2
122. Let u = --., du = -x dx.
2
_
1 _2x _ 2x
- -~e
- 1 _-2x
-~ + C
12~. f’(x) = [.½(e+~ + e-X)ax = ½(ex- e-x) + C,
f’(O) = C1 = 0
f(x) = [.½(e~ - e-X)dx = ½(ex + e-x) + C2
123. Let u = l+e2x,du = 2e2xdx.
!o~ dx =
I /1
f(O) = 1+ C2 = 1 ~ C~ = 0
f(x) = ½(ex + e-x)
+
= In(1 + e6) - ln2
130. f’(x) = sinx+e2x dx =-cosx+Te
+ C~
f’(0) = -1 4-1 ~ 4- C,1 = -~ ~ C1 = 1
124. Let u = 5 - M, du = -ex dx.
f’(x) = -cos x +1 _2x
2e + 1
~dx = -I25 - ex~
s(x) = l(- os 1+A.~
+ 1) dx
5 -- ex
1 _2x +x+C2
=-sinx+-~
f(0) = X1 + C2 -- ~" 1:=~ C2 = 0
= -ln(5 - e) + In 4
1 ~2x
f(x] = x - sin x + =~
In 4
131. (a)
Y
125. Let u = sin ~x, du = ~c cos ~cx dx.
ffl
!IS
Iit
lo~r/2eSin~X cos ~cx dx = -~ !~/2eSin~x(zc cos ~x) dx
~k
(0, ,,I
’)t sI~l
Ill! " / i / //--,-’, .........
’ i i.!//..’..’....-"5
"- / / ,1,,/./.,, ,,., .......
., 111.,......+~... .......
AO
Sl~l-: ’-7, ........................
(b)
dy = 2e_xl2 (0,1)
dx
’
126. Let u = sec 2x, du = 2 sec 2x tan 2x dx.
e~r/2
~.
[~/~e~°~x sec 2xtan 2x dx = 1 J,d~ e .... X{2sec 2xtan 2x)dx
~/~
2
= l[e~O z~-]~/~
2u
-~/~
=~[e-l-e-2]
y= 12e-Xl~ dx = -41e-xl2(-~ dx)
= -me-x/2 + C
(0,1): 1 =-4e° + C =-4+ C ~ C = 5
y = -4e-x/2 + 5
6
= "~[~ - -~TJ= 2e~
/
© 2010 Brooks/Cole, Cengage Learning
5211
Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
~32. (a)
~ /---4
y = Ixe-°’2"~2 dx = l~Ie-°’2x2(-O.4x) dx
-0.4
= -2.5e-°’2x2 + C
- --~1 e-°’2:~2 + C
0.4
-~ = -2.5e°+C =-2.5+C ~ C = I
y = -2.5e-°’2x2 + 1
x
137o Jo.,l xe dx, n : 12
Midpoint Rule: 92.1898
Trapezoidal Rule: 93.8371
Simpson’s Rule: 92.7385
Graphing utility: 92.7437
138o
2xe-Xdx, n = 12
Midpoint Rule: 1.1906
Trapezoidal Rule: 1.1827
Simpson’s Rule: 1.1880
Graphing utility: 1.18799
139. 0.0665 I~] e-°’°139(t-48)2dt
Graphing utility: 0.4772 = 47.72%
= e5 -1 ~ 147.413
120.3-°.3t dt 21= -
141).
-e-°’3x + 1 = -1
2
e_O.3x
: e-a _ e-b
1
2
-0.3x = ln-1 = -In2
2
ln2
x = ~ = 2.31 minutes
0.3
141o x(t) = Ae~1 + Be-k’; A, B, k > 0
(a) x’(t) = Ake~ - Bke-~ = 0
Aekt = Be-kt
B
A
135.
= -2e-3/2 + 2
= 1.554
3
t = 2~
-3
By the first Derivative Test, this is a minimum.
(b) x"(t)= Ak~e~ + Bk~e-~’
kz is the constant ofpropo~ionality.
@ 2010 Brooks/Cole, Cengage Learning
Section 5.4 Exponential Functions." Differentiation and Integration 521
142.
t
0
1
2
3
4
R
425
240
118
71
36
lnR
6.052
5.481
4.771
4.263
3.584
(a) In R = -0.6155t + 6.0609
R = e-0’6155t+6’0609 = 428.78e-°’6~55t
4~o
(b)
147.
[e’]~ >_
e:’ -1 > x ~ ex > 1 + xforx > 0
148. The graphs of f(x) : In x and g(x) : ex are mirror
images across the line y = x.
149.
-1
l~ et dt >_ l~ l dt
e-x = x ~ f(x) = x-e-x
f’(x) = 1+ e-~"
y(x,,) _
o
Xn+l-= Xn
f’(x,)
(C) I2R(t)dt =ao[4428"78e-°’6155’ dt
Xn -- e-Xn
1 + e-x"
xl =1
637.2 liters
f(x,) .~ 0.5379
X2 = X1 --_
143. f(x) = eX.Domain is (-~o, ~)and range is (0, c~). f is
continuous, increasing, one-to-one, and concave upwards
on its entire domain.
limex = 0andlimex = o~.
X--+--oo
144o Yes. f(x) = Cex, C a constant.
145. (a) Log Rule: (u = ex + 1)
(b) Substitution: (u = x2)
X3 = X2 --_
f(x2) ~ 0.5670
Z’(~)
X4 = X3
f(x3) ~ 0.5671
f’(x3)
Approximate the root off to be x ~ 0.567.
150. Area = 8_ = !_~e_Xdx
o
3
= -e-X3~
a
= -e-~ + e~
Let z = ca:
8
146. (a)
-1
8
-z = -1
3
-4
+ z2
3Z2 -- 8z - 3 = 0
-2
(b) When x increases without bound, 1Ix approaches
zero, and el/x approaches
Xn
1. Therefore, f(x)
approaches 2/(1 + 1) = 1.So, f(x)has a horizontal
asymptote at y = 1. As x approaches zero from the
right, 1Ix approaches ~o, e~/x approaches ~o and
(3z+l)(z-3) = 0
z = 3 ~ e~ =3~ a=ln3
1
e,~-- impossible
z = --1 ~ =
3
3
]
So, a = In 3.
f(x) approaches zero. As x approaches zero from
the left, 1Ix approaches -~o, el/x approaches zero,
and f(x) approaches 2. The limit does not exist
because the left limit does not equal the right limit.
Therefore, x = 0 is a nonremovable discontinuity.
© 2010 Brooks/Cole, Cengage Learning
522
Chapter 5
151. y -
Logarithmic, Exponential, and Other Transcendental Functions
a > O,b > O,L > 0
aL e-x] b
b
y
tt ~
y" = Oifae-x/b = 1 ~ b
L
y(b, In a], = - --(6’"‘,)/b
1+ ae
1+ a(1/a) 2
Therefore, the y-coordinate of the inflection point is L/2.
152. f(x) - lnx
x
(a) f’(x) - 1 -x2
In x= Owhenx = e.
On (0, e),f’(x) > 0 ~ f is increasing.
On (e, oo),f’(x) < 0 ~ f is decreasing.
(b) For e < A < B, youhave
lnA lnB
-- >
A B
BInA > AInB
In As > In Ba
AB > B"4.
(c) Because e < ~, from part (b) you have e
y
Section 5.5 Bases Other than e and Applcations
1. log2½ = log22-3 = -3
4. log‘,--1 = log,,1-1og,,a = -1
a
2. 10g27 9 = 10g27 272/3 = --2
3
5. (a)
23 =8
log2 8 = 3
3o log7 1 = 0
(b)
3-~ = _.1
3
log3 1
~ = -1
© 2010 Brooks/Cole, Cengage Learning